Solve x² + 4x + 1 = 0
The solutions of x² + 4x + 1 = 0 are x = -2 + √3 or x = -2 - √3, that is x ≈ -0.26794919 and x ≈ -3.7320508.
Change a, b or c (whole numbers, a not 0) and the exact solutions follow.
How to work it out
- 1
Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 1, b = 4, c = 1.
- 2
Work out the discriminant D = b² − 4ac. Its sign tells you how many real solutions there are.
- 3
D is positive, so there are two different real solutions.
- 4
The quadratic formula works for every quadratic:
- 5
Simplify (pull the square out of the root) to get the solutions:
- 6
As decimals: x ≈ -0.26794919 and x ≈ -3.7320508.
- 7
Check numerically: for x ≈ -0.26794919 the left side is -4.4408921e-16 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ x^2 + 4x + 1 = 0\ \Rightarrow\ x = -2 + \sqrt{3},\ -2 - \sqrt{3} \]
\[ x^2 + 4x + 1 = 0 \]
\[ D = 4^2 - 4\cdot1\cdot1 = 16 - 4 = 12 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm \sqrt{12}}{2} \]
\[ x = -2 + \sqrt{3}\quad\text{or}\quad x = -2 - \sqrt{3} \]The parabola y = x² + 4x + 1
| x | y |
|---|---|
| -5 | 6 |
| -4 | 1 |
| -3 | -2 |
| -2 | -3 |
| -1 | -2 |
| 0 | 1 |
| 1 | 6 |
Questions
What are the solutions of x² + 4x + 1 = 0?
The solutions of x² + 4x + 1 = 0 are x = -2 + √3 or x = -2 - √3, that is x ≈ -0.26794919 and x ≈ -3.7320508.
Does x² + 4x + 1 = 0 factor?
Not over the integers, because the discriminant 12 is not a perfect square. Use the quadratic formula.
What is the vertex of y = x² + 4x + 1?
(-2, -3). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = -4, product = c/a = 1.